Question

Solve the following simultaneous equation using Cramer’s rule given by $3x+y=1$ and $2x=11y+3$.
(a) x = $\dfrac{2}{5}$, y = $-\dfrac{7}{3}$
(b) x = $\dfrac{2}{5}$, y = $-\dfrac{1}{5}$
(c) x = $\dfrac{2}{5}$, y = $\dfrac{1}{5}$
(d) x = $\dfrac{2}{5}$, y = $\dfrac{7}{5}$

Answer 1

Hint: We will apply here the formulas $x=\dfrac{{{D}_{x}}}{D}$ and $y=\dfrac{{{D}_{y}}}{D}$. Where D is the determinant and ${{D}_{x}}$ and ${{D}_{y}}$ are the determinants for x and y variables. We will use the expansion of determination here which is done as by $\left| \left. \begin{matrix} a \\\ c \\\ \end{matrix}\,\,\begin{matrix} b \\\ d \\\ \end{matrix} \right| \right.$ into ad – bc.

Complete step-by-step answer:
The Cramer’s rule is a type of method which is used to find the values of x and y with the help of a determinant. This is a much easier method than elimination method.
Here we will have D as a determinant. Also we will use ${{D}_{x}}$ and ${{D}_{y}}$ as the determinants for x and y variables. Here for the variable x we will divide the determinant D by ${{D}_{x}}$ and for y we will divide determinant D by ${{D}_{y}}$.
The equation $2x=11y+3$ should be written as $2x-11y=3$.
Now, we will consider the equations that are given to us as $3x+y=1$ and $2x-11y=3$. Now we will convert these into determinant form. This can be done as below.
We will consider the matrix as $\left| \left. \begin{matrix} a \\\ c \\\ \end{matrix}\,\,\begin{matrix} b \\\ d \\\ \end{matrix} \right| \right.$. Now to insert the values of equation $3x+y=1$ into the matrix we will write a = 3 and b = 1. Thus we get $\left| \left. \begin{matrix} 3 \\\ c \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ d \\\ \end{matrix} \right| \right.$. Now we will put the value of the equation $2x-11y=3$ into the matrix. Thus, we will substitute c = 2 and d = - 11 into the matrix $\left| \left. \begin{matrix} 3 \\\ c \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ d \\\ \end{matrix} \right| \right.$. Therefore we get $\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.$. Now this will be treated as D. Thus we now have D = $\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.$.
Now, we will find the matrix for x. For that we will do the changes in the first column of the matrix D = $\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.$. Now look at the equations $3x+y=1$ and $2x-11y=3$. In the right side of the equations we have 1 and 3 numbers. So, we will substitute 1 and 3 in the first column of the matrix D = $\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.$. Therefore, the matrix changes into $\left| \left. \begin{matrix} 1 \\\ 3 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.$. Now, this will be regarded as ${{D}_{x}}$.
Similarly we will find the matrix for y. For that we will do the changes in the second column of the matrix D = $\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.$. In this matrix we will substitute 1 and 3 in the second column of the matrix D = $\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.$. Therefore, the matrix changes into $\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ 3 \\\ \end{matrix} \right| \right.$. Now, this will be regarded as ${{D}_{y}}$.
So now for finding the value of x we will divide the determinant D by ${{D}_{x}}$. This can be numerically written as $x=\dfrac{{{D}_{x}}}{D}$. After substituting the matrices into their particular positions we will get
$\begin{aligned} & x=\dfrac{{{D}_{x}}}{D} \\\ & \Rightarrow x=\dfrac{\left| \left. \begin{matrix} 1 \\\ 3 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.}{\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.} \\\ \end{aligned}$
Now we will evaluate the determinant which is done by $\left| \left. \begin{matrix} a \\\ c \\\ \end{matrix}\,\,\begin{matrix} b \\\ d \\\ \end{matrix} \right| \right.$ into ad – bc. Therefore, we get
$\begin{aligned} & x=\dfrac{\left| \left. \begin{matrix} 1 \\\ 3 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.}{\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.} \\\ & \Rightarrow x=\dfrac{1\times -11-1\times 3}{3\times -11-1\times 2} \\\ & \Rightarrow x=\dfrac{-11-3}{-33-2} \\\ & \Rightarrow x=\dfrac{-14}{-35} \\\ & \Rightarrow x=\dfrac{2}{5} \\\ \end{aligned}$
So now for finding the value of y we will divide the determinant D by ${{D}_{y}}$. This can be numerically written as $y=\dfrac{{{D}_{y}}}{D}$. After substituting the matrices into their particular positions we will get
$\begin{aligned} & y=\dfrac{{{D}_{y}}}{D} \\\ & \Rightarrow y=\dfrac{\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ 3 \\\ \end{matrix} \right| \right.}{\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.} \\\ \end{aligned}$
Now we will evaluate the determinant which is done by $\left| \left. \begin{matrix} a \\\ c \\\ \end{matrix}\,\,\begin{matrix} b \\\ d \\\ \end{matrix} \right| \right.$ into ad – bc. Therefore, we get
$\begin{aligned} & \Rightarrow y=\dfrac{\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ 3 \\\ \end{matrix} \right| \right.}{\left| \left. \begin{matrix} 3 \\\ 2 \\\ \end{matrix}\,\,\begin{matrix} 1 \\\ -11 \\\ \end{matrix} \right| \right.} \\\ & \Rightarrow y=\dfrac{3\times 3-1\times 2}{3\times -11-1\times 2} \\\ & \Rightarrow y=\dfrac{9-2}{-33-2} \\\ & \Rightarrow y=\dfrac{7}{-35} \\\ & \Rightarrow y=-\dfrac{1}{5} \\\ \end{aligned}$
Therefore the values of x = $\dfrac{2}{5}$ and the value of y = $-\dfrac{1}{5}$ .
Hence, the correct option is (b).

Note: This question can also be done by elimination or substitution method. But since this question specifically mentions that it should be solved by Cramer’s rule therefore the question is solved in this type. While expanding the determinant one mistakes in $\left| \left. \begin{matrix} a \\\ c \\\ \end{matrix}\,\,\begin{matrix} b \\\ d \\\ \end{matrix} \right| \right.$ into ad + bc which makes the answer wrong. So, to do it correctly apply $\left| \left. \begin{matrix} a \\\ c \\\ \end{matrix}\,\,\begin{matrix} b \\\ d \\\ \end{matrix} \right| \right.$ into ad – bc. The values of x and y can also be asked in points. If asked then we will write x = 0.4 and y = - 0.2. As we can see that the equation given to us is $2x=11y+3$. But while forming the determinant we should convert it into $2x-11y=3$ and then proceed. Solving with these important points we will be able to solve the question in a better and correct way.